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Can someone put up a tutorial on Integral Calculus?


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#1 Abhi_KS

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Posted 28 December 2013 - 03:40 AM

Can someone put up a tutorial on Integral Calculus just like there is one for differentiation? I need it.


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#2 EdEarl

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Posted 28 December 2013 - 04:18 AM

See https://www.khanacademy.org/
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#3 Abhi_KS

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Posted 28 December 2013 - 10:38 AM

Thanks, but I need more text oriented and Khan Academy teach in videos, which take eons to load on my slow internet.


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#4 John

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Posted 28 December 2013 - 01:14 PM

One good resource is Paul's Online Math Notes. The last couple of sections of Calculus I introduce integration, substitution (one of several techniques of integration), and applications of the integral. Calculus II then continues with more integration techniques and applications before getting into other calculus topics. There are also notes for Calculus III (which is multivariable calculus, though some of that material is introduced at the end of Calculus II) and Differential Equations. PDFs of the notes are available if you prefer those over the website, and Calculus I and II (but not III or DE) also come with practice problems and solutions.

There's also William Chen's lecture notes, which cover quite a few different mathematical subjects, including calculus. The notes include exercises, but I don't think worked solutions are available.

 

As for tutorials here, there has been some talk (though not really recently, that I know of) of adding more tutorials, but with the huge number of resources already available online for pretty much any math topic, adding more tutorials here probably isn't a priority.


Edited by John, 28 December 2013 - 01:14 PM.

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#5 Endercreeper01

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Posted 29 December 2013 - 02:05 AM

Do you want someone on science forums to explain it to you? Or do you want a website?


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#6 Abhi_KS

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Posted 29 December 2013 - 05:24 AM

Well, I would prefer it here, and more if it was just like the previous one.


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#7 Endercreeper01

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Posted 30 December 2013 - 04:33 AM

Here will be a tutorial on integral calculus. 

Part 1. Riemann Sums

An integral is the area under a curve. You can approximate an integral by adding together the area of a number of rectangles, each with a height of f(x) and a length of \Delta x. They each have an area that is equal to f(x)\Delta x. You take this sum from x = a to x = b, and since you are summing n rectangles, \Delta x will be the length from a to b, or a - b divided by the number of rectangles. This is written as \Delta x = \frac{a-b}{n} The sum of the areas of each rectangle will give you something close to the value of the integral. This is called a Riemann Sum. It is useful in approximations of integrals. As the rectangles have smaller and smaller lengths, the sum becomes more and more accurate, and you get a closer and closer approximation. This means that you have to use a limit. Therefore, the value of an integral is 

\lim_{n \to \infty} \sum_{i=1}^{n} f(x_i)\Delta x

You can imagine a function as being made up of an infinite number of these rectangles, each with an infinitesimal length of dx and height of f(x). An integral is written as

\int_a^b f(x) dx

which represents the infinite sum from a to b of the areas of each infinitesimal area.

Part 2. The Second Part of The Fundamental Theorem of Calculus

There is another formula for the integral, which was worked out by Issac Newton. Issac Newton Found out that integrating is the inverse function of differentiating. Because the area under a curve is evaluated from a to b, this means that the exact value of the integral is

\int_a^b f(x) dx = F(b) - F(a)

Where F(b) and F(a) are the antiderivatives of the function f(x) with respect to a and b.

This lets you know the exact area under a curve without working out a Riemann Sum.

 

That is the introduction to integration. In the next tutorial, you will learn about integration techniques.


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#8 Abhi_KS

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Posted 30 December 2013 - 06:11 AM

@Endercreeper01 : That was what I wanted. Thanks!


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#9 Endercreeper01

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Posted 30 December 2013 - 03:28 PM

Your welcome. Do you also want a tutorial for the rest of integral calculus?


Edited by Endercreeper01, 30 December 2013 - 10:51 PM.

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#10 Abhi_KS

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Posted 31 December 2013 - 05:15 AM

Yes, why not? This was exactly in the format and language which I wanted.


Edited by Abhi_KS, 31 December 2013 - 05:19 AM.

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#11 Endercreeper01

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Posted 31 December 2013 - 04:35 PM

Ok, so let's continue.

Part 3: Integral properties

Before we move on, you must know the properties of integrals.

\int_a^b f(x) + g(x) dx = \int_a^b f(x) dx + \int_a^b g(x) dx

\int_b^a f(x) dx = -\int_a^b f(x) dx

\int_a^b cf(x) dx = c\int_a^b f(x) dx where c is a constant

\int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx where c is a number.

Part 4: Antiderivatives

Antiderivatives are the opposite of derivatives. It's like being given f'(x) and trying to figure out f(x). Usually, they are written as \int f(x) dx, and this type of integral is called an improper integral. They are also written as F(x). You find antiderivatives by figuring out which function's derivative would be your original function, and that function would be the antiderivative. For example, if we wanted to find the antiderivative of sin(x), we know that the derivative of cos(x) is -sin(x), so therefore, the derivative of -cos(x) is sin(x). This means that the antiderivative of sin(x) is cos(x).

We can do this with much more functions. The antiderivatives of simple functions will be in the next section, and you will need these for more complex integration. This method of antidifferentiating is how mathematicians found these antiderivatives.

Part 5: Simple Antiderivatives

Here will be a list of simple antiderivatives. You will need to know these before we learn more complex integration.

\int x^n dx = \frac {x^{n+1}}{n+1} (except for n = -1)

\int e^x dx = e^x

\int a^x dx = \frac{a^x}{ln(a)}

\int ln(x) = xln(x) - x

\int \frac{1}{x} dx = ln|x|

\int sin(x) dx = -cos(x)

\int cos(x) dx = sin(x)

\int tan(x) dx = ln|sec(x)|

\int sec(x) dx = ln|sec(x) + tan(x)|

 

Part 6. U subsitution

U substitution is a useful integration technique for complex integrals.

For example, say you wanted to find \int sin(3x + 8) dx. You could easily solve his problem with u substitution. You would need to set a value to u, and then integrate with respect to u. In this example, we would set  u = 3x + 8, making this \int sin(u) dx. What we need to do now is make this integral with respect to u. If you imagine \frac{du}{dx} as being the ratio between 2 infinitesimals, you can multiply this by dx to get du. In this case, since u = 3x + 8, this means \frac{du}{dx} = 3 , and du=3dx. Because we need to take the integral with respect to u, we change dx to \frac{1}{3} du, since \frac{1}{3} du = dx. We bring the 1/3 out front, and this becomes \frac{1}{3} \int sin(u) du.

This gives you -\frac{1}{3} cos(u) + C. We now have to plug in the value of u, and because  u = 3x + 8, the integral works out to be -\frac{1}{3}cos(3x+8)+C. Don't forget the constant of integration, since you always need to add that.

The next lesson will be on u substitution practice.

Remember, if you need help with any of this, you can just ask our calculus forum.


Edited by Endercreeper01, 31 December 2013 - 09:53 PM.

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#12 imatfaal

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Posted 31 December 2013 - 09:40 PM


Ok, so let's continue.

...We now have to plug in the value of u, and because  u = 3x + 8, the integral works out to be \frac{1}{3}cos(3x+8).

The next lesson will be on u substitution practice.

Remember, if you need help with any of this, you can just ask our calculus forum.

 

surely the integral works out as -1/3 cos(3x+8) .  Also you need to add a constant - or explain why you should add one


Edited by imatfaal, 31 December 2013 - 09:42 PM.
horrid horrid editor

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#13 Endercreeper01

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Posted 1 January 2014 - 12:08 AM

surely the integral works out as -1/3 cos(3x+8) .  Also you need to add a constant - or explain why you should add one


The post has been edited. For some reason, the negative didn't show up. Also, I explained the constant of integration and why there should be one in the part about antiderivatives.
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#14 John

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Posted 1 January 2014 - 12:29 AM

I don't see it. Also, another slight edit: In the antiderivatives section, you refer to the antiderivative as an "improper" integral. This should be "indefinite" integral. I'm assuming you'll discuss improper integrals later.

Edited by John, 1 January 2014 - 12:29 AM.

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#15 Endercreeper01

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Posted 1 January 2014 - 03:32 PM

Because I am unable to edit the post now, I will add a short part about the constant of integration:

Part 4.5:The constant of integration

When you take antiderivatives, you need to add something called a constant of integration. During differentiation, you get rid of any constant terms, so when you antidifferentiate, you must add something called a constant of integration to include any constant terms lost when differentiating. This is only for indefinite integrals. If you have a definite integral, you do not need to include the constant of integration, since when you work out F(b) - F(a), F(b) and F(a) have the same constant of integration. Because of this, the two constants of integration would cancel out when you are subtracting, and this means there is no constant of integration in definite integrals.


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#16 studiot

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Posted 1 January 2014 - 04:51 PM

 

When you take antiderivatives, you need to add something called a constant of integration. During differentiation, you get rid of any constant terms, so when you antidifferentiate, you must add something called a constant of integration to include any constant terms lost when differentiating. This is only for indefinite integrals. If you have a definite integral, you do not need to include the constant of integration, since when you work out 40e6c7792bd33eb7b8b66c9d95651c4a-1.png, F(b) and F(a) have the same constant of integration. Because of this, the two constants of integration would cancel out when you are subtracting, and this means there is no constant of integration in definite integrals.

 

 

 

There is more to it than this.

 

Others have already said that you should be comparing the indefinite integral and the definite integral.

 

The anti-derivative is something else again.

 

The definite integral is a pure number it is not a function.

 

The indefinite integral is a function.

 

I suggest accepting these for now until the significance of this distinction really becomes apparent when you proceed to the applications of calculus.

 

And well done (+1) to endercreeper for being prepared to put up the work.


Edited by studiot, 1 January 2014 - 04:54 PM.

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#17 Endercreeper01

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Posted 1 January 2014 - 10:28 PM

Oh yes, thanks for reminding me. I will make sure to mention that when we get to the applications of integral calculus.

I have decided to skip the practice problems. Upon request, I will provide them, but I have decided to skip them for now.

Part 7. Integration By Parts

Integration by parts is like the product rule for integrals. Let's define f(x) as being f(x) = uv, where u and v are both functions of x. We know that the derivative of this function will be f'(x) = (uv)' = u'v + v'u. Because of this, we can integrate u'v + v'u to get the original function, i.e \int {u'v + v'u} dx = uv. Integrals are distributive, so we can write the previous equation as uv = \int u'v dx + \int v'u dx. In order to integrate uv', you have to subtract the integration of vu' from the equation, and this would have to be from both sides of the equation. We find that \int uv' dx = uv - \int vu' dx. Using the fact that u' dx = du and  v' dx = dv, we obtain the equation for integration by parts:

\int u dv = uv - \int v du

For example, let's integrate 2x ln(x).  We can set the value of dv to 2x dx, and therefore, v = x^2. From here, you can see that u = ln(x), since that is the only other part of the function. We now need to plug the value of u and v into the equation. Because u = ln(x) and v = x2, we find that \int 2x ln(x) dx = x^2 ln(x) - \frac{x^2}{2}, since du = 1/x dx and \frac{x^2}{x}=x. Because of that, the integral turns into \int x dx, which turns out to be x2/2.


Edited by Endercreeper01, 1 January 2014 - 10:30 PM.

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#18 Abhi_KS

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Posted 3 January 2014 - 12:06 PM

Endercreeper01 rulez!

 

I think now this thread can be counted in Tutorial Section.


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#19 Endercreeper01

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Posted 3 January 2014 - 05:50 PM

Thanks for the support!
Now, we will discuss improper integrals.
Part 8. Improper integrals

An improper integral is when one or two limit(s) of a definite integral is infinity. For example, \int_{-\infty}^{5} \frac{1}{e^x} dx and \int_0^\infty x^{3x} dx are both improper integrals.

Of course, you can't work this out by just plugging in infinity. For improper integrals, you have to use a limit. You could write the above integral as \lim_{n \to \infty} \int_0^n x^{3x} dx., and this would mean you would have to take the limit as n approaches infinity or negative infinity for that function. Improper integrals are categorized 2 ways: divergent and convergent. It is convergent if the integral exists and is not infinite, and divergent if it is infinite or it doesn't exist. For instance, let's work out the integral \int_0^\infty \frac{1}{x}. We know the antiderivative of this function is ln(x). We also know that ln(x) approaches infinity as x approaches infinity. Therefore, this integral is divergent.

That concludes the tutorial on calculating integrals. Next, you will learn about the applications of integrals.

Part 9. Average Function Value 

We know that in order to calculate the average of a set of numbers, you must add together all of the parts of the set and divide by the number of numbers in the set. If you had an infinite number of numbers in a set, such as function values from a to b, you would have to use a different formula. The equation for the average function value from a to b is that f_{avg}=\frac{\int_a^b f(x) dx}{b-a}. This is because we know that in a Riemann sum, \Delta x = \frac{b-a}{n}, and when n is infinity, we represent it by dx (an infinitesimal). Dividing by b-a in the denominator gets rid of the b-a in the dx, and so, it is the infinite sum of the infinitesimals of \frac{f(x)}{\infty}. For example, let's try \frac{\int_1^5 x^2 dx}{4}. The denominator is 4, since the difference between the top and bottom limit is 4, and because the antiderivative is \frac{x^3}{3}, we find the difference between this at x = 5 and x = 1, then divide by 4. Working this out, we find that f_{avg} = 10 + 1/3.

 

Next, you will learn about calculating arc length.

Remember, if you need any help with this, you can ask our calculus forum.


Edited by Endercreeper01, 3 January 2014 - 10:12 PM.

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#20 Endercreeper01

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Posted 5 January 2014 - 05:18 PM

Part 10. Arc Length of A Curve

The arc length of a curve from a to b is how long the function graph from a to b would be. To approximate the arc length, we can break the function up into n lines, each with length \Delta s. An estimate for the arc length would be to write it as the sum of the lengths of these lines, I.e s \approx \sum_{i=1}^{n} \Delta s_i. As we have more and more lines, the approximation becomes more and more accurate, and so we take the limit as n approaches infinity of the above sum. We find that s = \lim_{n \to \infty} \sum_{i=1}^{n} \Delta s_i. We know that an integral is an infinite sum, so we have to integrate over all of the infinitesimal lengths. We now obtain the equation for arc length of a curve:

s=\int_a^b ds

Because we know that ds=\sqrt {dx^2+dy^2} and that ds is also \sqrt {1 + \left( \frac {dy} {dx} \right)^2} dx we can rewrite the previous equation as

s=\int_a^b \sqrt {1 + \left( \frac {dy} {dx} \right) ^2} dx

which gives us the equation for arc length. You can also define x = f(y) and use the equation:

s=\int_a^b \sqrt {1 + \left( \frac {dx} {dy} \right) ^2} dy

as this would also work. 

For example, we will work out the arc length from 1 to 3 of y=\frac{2\sqrt {x^3}}{3}.

The derivative of this function will be \sqrt x. When we square this derivative, we obtain x.  

The integral now becomes \int_1^3 \sqrt {1+x} dx. The substitution will be u = 1+x, and so du = dx. This integral now becomes \int_1^3 \sqrt u du. The antiderivative of this function is \frac{2}{3}u^{3/2}, which is equal to \frac{2}{3}(1+x)^{3/2}. In order to get the answer, we must work this out at x=3, then subtract by the antiderivative value at x=1. 

Our final answer ends up as \int_1^3 \sqrt {1+x} dx = \frac{16}{3} - \frac{4\sqrt 2}{3}.

 

The next tutorial will be on the area between two curves.

Remember, if you need any help with this, you can just ask our calculus forum.


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